Cracking The Ap Calculus ab Exam 2018

We need to find an expression for the distance from the point P to the point (2, 1) and then
minimize the distance. If we call the coordinates of P (x, y), then we can find the distance to (2,

1. using the distance formula: D^2 = (x − 2)^2 + (y − 1)^2 . Next, just as we did in sample problem
   4 (this page), we can let L = D^2 and minimize L: L = (x − 2)^2 + (y − 1)^2 = x^2 − 4x + 4 + y^2 − 2y

Because x^2 + y^2 = 1, we can substitute for y to get L = x^2 − 4x + 4 + (1 − x^2 ) − 2 + 1,

which    simplifies  to L    =   −4x     +   6   −   2 .     Next,   we  take    the     derivative:     =  

    =    .  Next,   we  set the derivative  equal   to  zero:

−4  +       =   0.  The best    way to  solve   this    is  to  move    the 4   to  the other   side    of  the equals

sign    and cross-multiply.